How do you solve using the completing the square method #x^2-6x+5=0#?

3 Answers
Jul 15, 2016

Answer:

#color(blue)(x=5)# or #color(blue)(x=1)#

Explanation:

Remember in squaring a general binomial
#color(white)("XXX")(a+b)^2=a^2+2ab+b^2#

Given: #x^2-6x+5=0#

If #x^2# and #-6x# are the first two terms of the expansion of a squared binomial:
#color(white)("XXX")a=1#, and
#color(white)("XXX")b=-3#

So the third term must be #b^2=(-3)^2=9#

To complete the square we add a #9# to the first two terms (but to keep everything balanced we will need to subtract it again:
#color(white)("XXX")x^2-6xcolor(red)(+9)+5color(red)(-9)=0#

Now we can write the first 3 terms as a squared binomial:
#color(white)("XXX")(x-3)^2+5-9=0#

Simplifying #(5-9)=-4# and adding #4# to both sides gives
#color(white)("XXX")(x-3)^2=4#

Taking the square root of both sides
#color(white)("XXX")x-3=+-2#

Then add #3# to both sides
#color(white)("XXX")x=3+-2#

#rArr x=5 or x=1#

Jul 15, 2016

Answer:

#x = 5, " or " x =1#

Explanation:

It seems strange to want to use completing the square, because this quadratic trinomial factorises to give #(x-5)(x-1) =0#

But let's proceed... Completing the square is based on the fact that
the square of a binomial gives a standard answer

#(x -8)^2 = x^2 -color(blue)(16)x +color(magenta)64#

There is ALWAYS a relationship between the #color(blue)16 and color(magenta)64#

This is: #color(blue)(16 div 2)" and then squared gives "color(magenta)64 " "8^2 =64#

In #x^2 -6x + 5 = 0 " the relationship is not correct"#

This means that 5 is not the correct constant, move it to the RHS

#x^2 -color(blue)(6)x color(magenta)(+9)= -5color(magenta)(+9)" add the correct constant to both sides"#

#(x-3)^2 = 4 " now the LHS is a perfect square"#

#x-3 = +-sqrt4" square root both sides"#

#x = 2+3, " or " x = -2+3#

#x = 5, " or " x =1#

Jul 15, 2016

Another example of method. Have a look at:

https://socratic.org/s/awchNPwZ